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3 years ago

Elements in Group VIIIA (also known as Group 18, or the noble gases) have

Physics

1 answer:

Sonbull [250]3 years ago

3 0

The answer is C (the same number of valence electrons)

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A boat is able to move through still water at 20 m/s. It makes a round trip to a town 3.0 km upstream. If the river flows at 5m/

Salsk061 [2.6K]

Answer:

Option e) 320 s

Explanation:

Here, distance = 3.0 km = 3000 m

The velocity of boat when it is going upstream;

Upstream velocity = velocity of boat in still water - velocity of river flow

So, Upstream velocity $=20m/s-5m/s=15m/s$

So,Time to go upstream

$(t_{1}) =\frac{Distance}{Velocity}=\frac{3000m}{15m/s} =200 s$

The velocity of boat when it is going downstream;

Downstream velocity = velocity of boat in still water + velocity of river flow

So, Downstream velocity $=20m/s+5m/s=25m/s$

So,Time to go downstream

$(t_{2}) =\frac{Distance}{Velocity}=\frac{3000m}{25m/s} =120 s$

So, total time (t) = $t_{1}+t_{2}=200s+120 s=320s$

Option E is the correct answer.

3 0

3 years ago

Read 2 more answers

How many species go extinct every day??

Law Incorporation [45]

Approximately 150-200 species.

5 0

3 years ago

How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.

nata0808 [166]

Hello!

Answer:
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement.

We are going to use the following formula:

$W= F . Cos \alpha . D$

Where:

$F=214 N$
$\alpha =0$ º
$D= 37m$

Then, by substituting we have:

$W=214N . Cos (0).37m= 7918 N.m=7918 J$

8 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago

Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem

Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

7 0

3 years ago